Linear Algebra/Describing the Solution Set/Solutions

Solutions
{{TextBox|1=  Solve the system of equations.
 * ? Problem 16:



\begin{array}{*{2}{rc}r} ax &+  &y  &=  &a^2  \\ x &+  &ay &=  &1 \end{array} $$

For what values of $$a$$ does the system fail to have solutions, and for what values of $$a$$ are there infinitely many solutions?  Answer the above question for the system.



\begin{array}{*{2}{rc}r} ax &+  &y  &=  &a^3  \\ x &+  &ay &=  &1 \end{array} $$

 (USSR Olympiad #174)
 * Answer:

This is how the answer was given in the cited source.

 Formal solution of the system yields



x=\frac{a^3-1}{a^2-1} \qquad y=\frac{-a^2+a}{a^2-1}. $$

If $$a+1\neq 0$$ and $$a-1\neq 0$$, then the system has the single solution



x=\frac{a^2+a+1}{a+1} \qquad y=\frac{-a}{a+1}. $$

If $$a=-1$$, or if $$a=+1$$, then the formulas are meaningless; in the first instance we arrive at the system



\left\{ \begin{array}{*{2}{rc}r} -x &+ &y  &=  &1 \\ x &- &y  &=  &1 \end{array}\right. $$

which is a contradictory system. In the second instance we have



\left\{ \begin{array}{*{2}{rc}r} x &+ &y  &=  &1 \\ x &+ &y  &=  &1 \end{array}\right. $$

which has an infinite number of solutions (for example, for $$x$$ arbitrary, $$y=1-x$$).  Solution of the system yields



x=\frac{a^4-1}{a^2-1} \qquad y=\frac{-a^3+a}{a^2-1}. $$

Here, is $$a^2-1\neq 0$$, the system has the single solution $$x=a^2+1$$, $$y=-a$$. For $$a=-1$$ and $$a=1$$, we obtain the systems



\left\{ \begin{array}{*{2}{rc}r} -x &+ &y  &=  &-1 \\ x &- &y  &=  &1 \end{array}\right. \qquad \left\{ \begin{array}{*{2}{rc}r} x &+ &y  &=  &1 \\ x &+ &y  &=  &1 \end{array}\right. $$

both of which have an infinite number of solutions.  }} {{TextBox|1= In air a gold-surfaced sphere weighs $$ 7588 $$ grams. It is known that it may contain one or more of the metals aluminum, copper, silver, or lead. When weighed successively under standard conditions in water, benzene, alcohol, and glycerine its respective weights are $$ 6588 $$, $$ 6688 $$, $$ 6778 $$, and $$ 6328 $$ grams. How much, if any, of the forenamed metals does it contain if the specific gravities of the designated substances are taken to be as follows? {{harv|Duncan|Quelch|1952}}
 * ? Problem 17:
 * Answer:

This is how the answer was given in the cited source.

Let $$ u $$, $$ v $$, $$ x $$, $$ y $$, $$ z $$ be the volumes in $$ {\rm cm}^3 $$ of Al, Cu, Pb, Ag, and Au, respectively, contained in the sphere, which we assume to be not hollow. Since the loss of weight in water (specific gravity $$ 1.00 $$) is $$ 1000 $$ grams, the volume of the sphere is $$ 1000\mbox{ cm}^3 $$. Then the data, some of which is superfluous, though consistent, leads to only $$ 2 $$ independent equations, one relating volumes and the other, weights.



\begin{array}{*{5}{rc}r} u &+  &v    &+  &x     &+  &y     &+  &z     &=  &1000  \\ 2.7u &+  &8.9v &+  &11.3x &+  &10.5y &+  &19.3z &=  &7558 \end{array} $$

Clearly the sphere must contain some aluminum to bring its mean specific gravity below the specific gravities of all the other metals. There is no unique result to this part of the problem, for the amounts of three metals may be chosen arbitrarily, provided that the choices will not result in negative amounts of any metal.

If the ball contains only aluminum and gold, there are $$ 294.5\mbox{ cm}^3 $$ of gold and $$ 705.5\mbox{ cm}^3 $$ of aluminum. Another possibility is $$ 124.7\mbox{ cm}^3 $$ each of Cu, Au, Pb, and Ag and $$ 501.2\mbox{ cm}^3 $$ of Al. }}